generating an equation for a problem

**clemkonan** · March 6th 2009, 05:27 PM

I am trying to help my grade 9 daughter solve this problem

Man has e mail about concert
Man sends email to 10 friends on day 1 so 10 people know about event
Each friend send e mail to 2 friends on day 2 so 20 people knows about event
Each of these 20 people sends email to 2 people on day 3 so 40 knows about event

Assume there are 5000 people in the town with e-mail and assuming mail only gets forwarded to people who have not already received it , how many days will it take before all 5000 people know about the concert.

Please kindly show the steps if possible.

**ADARSH** · March 6th 2009, 09:05 PM

Originally Posted by clemkonan

I am trying to help my grade 9 daughter solve this problem

Man has e mail about concert
Man sends email to 10 friends on day 1 so 10 people know about event
Each friend send e mail to 2 friends on day 2 so 20 people knows about event
Each of these 20 people sends email to 2 people on day 3 so 40 knows about event

Assume there are 5000 people in the town with e-mail and assuming mail only gets forwarded to people who have not already received it , how many days will it take before all 5000 people know about the concert.

Please kindly show the steps if possible.

-Initially 1 person knows about it
-Now 10 people know about it

Series comes out to be

$1,10,20,40,80,160,....$

Leave the first term (ie; leave 1 ) series now is

$10,20,40,80,..$

We need to find the condition "just" when

$1+10+20+40+80+...nth~term \ge 5000$

$10+20+40+80+...nth~term \ge 4999$

Now this is a Geometric Progression

The sum of n terms of a GP
a, ar, ar^2, ar^3,....ar^n

is given by

$\frac{a(r^n -1)}{r-1}$

Using this we get

$Sum = \frac{10(2^n-1)}{2-1} \ge 4999$

$10(2^n-1)\ge 4999$

$<br /> (2^n-1) \ge \frac{4999}{10}$

$2^n \ge \frac{5001}{10}$

For n= 9 LHS = 512

Hence on the 9th day this would be completely spread

**clemkonan** · March 7th 2009, 04:34 PM

In evaluating 2^n -1 the second last equation in your response , would the net effect not be adding +1 to the right hand side so that 4999 becomes 5000?

Or stated otherwise can you show how you simplified 2^n-1 so that you had 2^n isolated on the left hand side,

Otherwise I follow and sincere thanks

**ADARSH** · March 7th 2009, 08:22 PM

Originally Posted by ADARSH

$<br /> (2^n-1) \ge \frac{4999}{10}$

There was a great typo here though the answer remains same, Sorry for it
Add 1 on both sides

$<br /> (2^n) \ge \frac{4999}{10}+1$

$<br /> (2^n) \ge \frac{4999+10}{10}$

$2^n \ge 500.9$ ..........5001/10 is replaced by 5009/10

For n= 9 LHS = 512

Hence on the 9th day this would be completely spread

Red

**clemkonan** · March 8th 2009, 12:05 PM

Thanks ADARSH my goal was to demonstrate to the kids that the computer is a great resource and is good for much more than Vibe or Youtube. Yout\r time is greatly appreciated.
Clemkonan

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